i) Selection of operating point

Assume that $R_{C}\geq\geq R_{L}$

$R_{C}=\frac{1}{10}R_{L}=\frac{120K}{10}=12 k\Omega \ \ \ .....(i)$

Now select the $V_{CE}=3 V and V_{E}=5\times V_{BE}=5\times 0.7=3.5$

$V_{RC}=V_{CC}-V_{CE}-V_{E}$

$R_{C}=\frac{V_{RC}}{I_{C}} I_{C}=\frac{17.5V}{12K}=15 mA$

Operating point will be (3 V.1.5mA) ...........(ii)

ii)Selection of transistor

Select transistor with $V_{CE}\ge$ 3 V and $P_{Dmax}=3\times 1.5mA=4.5mW$

Select transistor 2N5835 with following specifications . . . ....(iii)

iii)Selection of $R_{E}$

The emitter resistor

$R_{E}=\frac{V_{RE}}{I_{E}}=\frac{3}{1.5mA}=2 k\Omega \ \ \ \ ......(iv)$

iv)Selection of R_{1} and R_{2}

For better stability, the current flowing through resistor $R_{1}$ is approximately greater than 10 times base current or current through $R_{2} is (1/10)^{th}$ of $1_{C}$

$I_{2}=\frac{I_{C}}{10}=\frac{1.5mA}{10}=0.15 mA$

The voltage at base of transistor is

$V_{B}=V_{BE}+V_{RE}$

=0.7+3.5=4.2V

$R_{2}=\frac{V_{B}}{I_{2}}$=$\frac{4.2}{0.15mA}=28k\Omega \ \ \ \ .....(v)$

and $R_{1}=\frac{V_{CC}-V_{B}}{I}$=$\frac{24-4.2}{210\mu A}94.285k\Omega=100k\Omega \ \ \ \ .....(vi)$ Since $I_{f}=I_{B}+I_{2}$ but $I_{B}=I_{C}/h_{FE}=15 mA/25=60\mu A$ very less

$I_{1}=60 \mu A+0.15mA=210 \mu A$

$R_{B}$ is parallel combination of $R_{1}$ and R_{2}

$R_{B}=\frac{R_{1}R_{2}}{R_{1}+R_{2}}=\frac{100K\times 28K}{100K+28K}$

$R_{B}=21.87k\Omega=22k\Omega \ \ \ ....(vii)$

v) Selection of $C_{E}$

The capacitive reactance of bypass resistor is

$X_{CE}=\frac{h_{ie}}{1+h_{fe}}$

But $h_{fe}=\frac{26mV}{I_{C}}(1+h_{fe})=\frac{26mV}{1.5mA}(1+25)=450\Omega=1k\Omega$

$X_{CE}=\frac{450}{26}=17.33\Omega$

$C_{E}=\frac{1}{2\pi \times 100\times 17\Omega}=93.6\mu F=100\mu F \ \ \ .....(viii)$

vi) Selection of $C_{1} and C_{2}$

The coupling capacitor $C_{1}$ is given by,

$C_{1}=\frac{1}{2\Pi X_{C1}f_{0}}$

But $X_{C1}=\frac{Z_{i}}{10}=\frac{R_{1}||R_{2}||h_{fe}}{10}$

=$\frac{22k||450}{10}=\frac{440.97}{10}=44\Omega$

$C_{1}=\frac{1}{2\pi \times 100\times 44}=36\mu F=36 \mu F \ \ \ ....(ix)$

And $C_{2}=\frac{10}{2\pi R_{L}f_{0}}=\frac{10}{2\pi \times 12k\times 100}$

$C_{2}=0.132\mu F=0.1\mu F \ \ \ \ ....(x)$